Kirchhoff’s first rule – problems and solutions

1. Based on figure as shown below, if I1 = 5A, I2 = 3 A, I3 = 6 A, then I4 = ………

Kirchhoff’s first rule – problems and solutions 1Solution :

Kirchhoff’s first rule states that the sums of all currents entering the junction must equal the sum of all currents leaving the junction.

I1 and I2 are entering the junction, whereas I3 and I4 are leaving.

Apply Kirchhoff’s first rule :

I1 + I2 = I3 + I4

5 + 3 = 6 + I4

8 = 6 + I4

I4 = 8 – 6

I4 = 2 Ampere

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2. If I1 = I3 then I4 =……..

Kirchhoff’s first rule – problems and solutions 2Known :

I1 = I3 = 2 Ampere

I2 = 3 Ampere

I5 = 4 Ampere

Wanted: I4

Solution :

I1 and I4 are entering the junction, whereas I2, I3, and I5 are leaving.

Apply Kirchhoff’s first rule :

I1 + I4 = I2 + I3 + I5

2 + I4 = 3 + 2 + 4

2 + I4 = 9

I4 = 9 – 2

I4 = 7 Ampere

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3. I1 = …….

Solution :Kirchhoff’s first rule – problems and solutions 3

I am entering the junction, whereas I1, I2, and I3 are leaving.

Apply Kirchhoff’s first rule :

I = I1 + I2 + I3

7 A = I1 + 1 A + 4 A

7 A = I1 + 5 A

7 A – 5 A = I1

I1 = 2 Ampere

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4. I2 = ………

Wanted: I2Kirchhoff’s first rule – problems and solutions 4

Solution :

I1 is entering the junction, whereas I2, I3, and I4 are leaving.

Apply Kirchhoff’s first rule :

I1 = I2 + I3 + I4

6 = I2 + 3 + 1

6 = I2 + 4

I2 = 6 – 4

I2 = 2 Ampere

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5. If I2 = 1/4 I1 , then I3 = ………

Known :Kirchhoff’s first rule – problems and solutions 5

I1 = 600 milliAmpere

Wanted: I3

Solution :

I2 = 1/4 I1 = 1/4 (600 mA) = 150 mA

I1 is entering the junction, whereas I2 and I3 are leaving.

Apply Kirchhoff’s first rule :

I1 = I2 + I3

600 mA = 150 mA + I3

I3 = 600 mA – 150 mA

I3 = 450 mA

6. Based on figure below, find I!

Solution :

Apply Kirchhoff‘s first rule :

3 + 5 + 2 = 7 + IKirchhoff’s first rule – problems and solutions 10

10 = 7 + I

10 – 7 = I

I = 3 Ampere

7. Based on figure below, if I1 = 4 A, I2 = 3 A, I4 = 7 A, I5 = 4 A, determine I3.

Solution :

Electric current entering the junction = I1 + I3 = 4 A + I3

Electric current leaving the junction = I2 + I4 + I5 = 3 A + 7 A + 4 A = 14 AKirchhoff’s first rule – problems and solutions 11

14 A = 4 A + I3

I3 = 14 A – 4 A

I3 = 10 A